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AttributeError: 'str' object has no attribute 'keys' error in getting json object keys

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1

I'm trying to convert a json object to csv file. When I pass the obj and print it, it works perfectly. But when I continue from line #Problem, I get an AttributeError: 'str' object has no attribute 'keys'.
I pass the exact same object. How can I fix this error and let it run properly?

import json
import csv
import os

def flattenjson( b, delim ):
 val = 
 for i in b.keys():
 if isinstance( b[i], dict ):
 get = flattenjson( b[i], delim )
 for j in get.keys():
 val[ i + delim + j ] = get[j]
 else:
 val[i] = b[i]
 return val

jjj = 
 "pk": 22,
 "model": "auth.permission",
 "fields": 
 "codename": "add_message",
 "name": "Can add message",
 "content_type": 8
 
 

print(flattenjson(jjj , "__" ))
#Problem
input = map( lambda x: flattenjson( x, "__" ), jjj )

columns = [ x for row in input for x in row.keys() ]
columns = list( set( columns ) )

with open( 'sad.csv', 'wb' ) as out_file:
 csv_w = csv.writer( out_file )
 csv_w.writerow( columns )

 for i_r in input:
 csv_w.writerow( map( lambda x: i_r.get( x, "" ), columns ) )

python json csv export-to-csv

share|improve this question

asked Mar 8 at 4:36

SadafSadaf

62

add a comment | 

1

I'm trying to convert a json object to csv file. When I pass the obj and print it, it works perfectly. But when I continue from line #Problem, I get an AttributeError: 'str' object has no attribute 'keys'.
I pass the exact same object. How can I fix this error and let it run properly?

import json
import csv
import os

def flattenjson( b, delim ):
 val = 
 for i in b.keys():
 if isinstance( b[i], dict ):
 get = flattenjson( b[i], delim )
 for j in get.keys():
 val[ i + delim + j ] = get[j]
 else:
 val[i] = b[i]
 return val

jjj = 
 "pk": 22,
 "model": "auth.permission",
 "fields": 
 "codename": "add_message",
 "name": "Can add message",
 "content_type": 8
 
 

print(flattenjson(jjj , "__" ))
#Problem
input = map( lambda x: flattenjson( x, "__" ), jjj )

columns = [ x for row in input for x in row.keys() ]
columns = list( set( columns ) )

with open( 'sad.csv', 'wb' ) as out_file:
 csv_w = csv.writer( out_file )
 csv_w.writerow( columns )

 for i_r in input:
 csv_w.writerow( map( lambda x: i_r.get( x, "" ), columns ) )

python json csv export-to-csv

share|improve this question

asked Mar 8 at 4:36

SadafSadaf

62

add a comment | 

I'm trying to convert a json object to csv file. When I pass the obj and print it, it works perfectly. But when I continue from line #Problem, I get an AttributeError: 'str' object has no attribute 'keys'.
I pass the exact same object. How can I fix this error and let it run properly?

import json
import csv
import os

def flattenjson( b, delim ):
 val = 
 for i in b.keys():
 if isinstance( b[i], dict ):
 get = flattenjson( b[i], delim )
 for j in get.keys():
 val[ i + delim + j ] = get[j]
 else:
 val[i] = b[i]
 return val

jjj = 
 "pk": 22,
 "model": "auth.permission",
 "fields": 
 "codename": "add_message",
 "name": "Can add message",
 "content_type": 8
 
 

print(flattenjson(jjj , "__" ))
#Problem
input = map( lambda x: flattenjson( x, "__" ), jjj )

columns = [ x for row in input for x in row.keys() ]
columns = list( set( columns ) )

with open( 'sad.csv', 'wb' ) as out_file:
 csv_w = csv.writer( out_file )
 csv_w.writerow( columns )

 for i_r in input:
 csv_w.writerow( map( lambda x: i_r.get( x, "" ), columns ) )

python json csv export-to-csv

share|improve this question

asked Mar 8 at 4:36

SadafSadaf

62

import json
import csv
import os

def flattenjson( b, delim ):
 val = 
 for i in b.keys():
 if isinstance( b[i], dict ):
 get = flattenjson( b[i], delim )
 for j in get.keys():
 val[ i + delim + j ] = get[j]
 else:
 val[i] = b[i]
 return val

jjj = 
 "pk": 22,
 "model": "auth.permission",
 "fields": 
 "codename": "add_message",
 "name": "Can add message",
 "content_type": 8
 
 

print(flattenjson(jjj , "__" ))
#Problem
input = map( lambda x: flattenjson( x, "__" ), jjj )

columns = [ x for row in input for x in row.keys() ]
columns = list( set( columns ) )

with open( 'sad.csv', 'wb' ) as out_file:
 csv_w = csv.writer( out_file )
 csv_w.writerow( columns )

 for i_r in input:
 csv_w.writerow( map( lambda x: i_r.get( x, "" ), columns ) )

share|improve this question

asked Mar 8 at 4:36

SadafSadaf

62

share|improve this question

asked Mar 8 at 4:36

SadafSadaf

62

asked Mar 8 at 4:36

SadafSadaf

62

asked Mar 8 at 4:36

SadafSadaf

62

1 Answer
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0

while iterating in your lambda function you're giving the only key which is considered as the string

So here is the following solution that might work for you.

Edited the code instead of passing the only key as an argument I'm passing a new dict.

input = map( lambda x: flattenjson(x: jjj[x], "__" ), jjj )

share|improve this answer

answered Mar 8 at 4:51

Shubham SrivastavaShubham Srivastava

380416

add a comment | 

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0

while iterating in your lambda function you're giving the only key which is considered as the string

So here is the following solution that might work for you.

Edited the code instead of passing the only key as an argument I'm passing a new dict.

input = map( lambda x: flattenjson(x: jjj[x], "__" ), jjj )

share|improve this answer

answered Mar 8 at 4:51

Shubham SrivastavaShubham Srivastava

380416

add a comment | 

0

while iterating in your lambda function you're giving the only key which is considered as the string

So here is the following solution that might work for you.

Edited the code instead of passing the only key as an argument I'm passing a new dict.

input = map( lambda x: flattenjson(x: jjj[x], "__" ), jjj )

share|improve this answer

answered Mar 8 at 4:51

Shubham SrivastavaShubham Srivastava

380416

add a comment | 

while iterating in your lambda function you're giving the only key which is considered as the string

So here is the following solution that might work for you.

Edited the code instead of passing the only key as an argument I'm passing a new dict.

input = map( lambda x: flattenjson(x: jjj[x], "__" ), jjj )

share|improve this answer

answered Mar 8 at 4:51

Shubham SrivastavaShubham Srivastava

380416

input = map( lambda x: flattenjson(x: jjj[x], "__" ), jjj )

share|improve this answer

answered Mar 8 at 4:51

Shubham SrivastavaShubham Srivastava

380416

answered Mar 8 at 4:51

Shubham SrivastavaShubham Srivastava

380416

answered Mar 8 at 4:51

Shubham SrivastavaShubham Srivastava

380416